A cart is propelled over an xy plane with acceleration...

A cart is propelled over an xy plane with acceleration components ax = 4.0 m/s^2 and ay = -2.0 m/s^2. Its initial velocity has components v0x = 8.0 m/s and v0y = 12 m/s. Assuming the velocity of the cart as the greatest y-coordinate is v = (32 m/s)i, answer parts a, b, c, and d below. Show all work and give explanation. Be sure to use proper vector notation whenever necessary. a) In unit-vector notation, what is the velocity of the cart when it reaches its greatest y-coordinate? b) Obtain the formula for speed as a function of time, t. c) At what time does the cart move the fastest? d) Estimate the fastest speed.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 So here it is given that a cart is propelled over a xy plane with acceleration a x which is 4 meters per second square and and acceleration along y direction a y equal to negative 2 meters per second square rate the initial velocity is v0 x which is or v0 x which is 8 meters per second and v0 y is 12 meters per second right now you have to tell what is the velocity of the cart when it is it has reached the greatest y coordinate right now here we have got y to be equal to y max right now here if you see we have got negative acceleration right that means the velocity of this uh cart along the y direction is going to get reduced right so that means the maximum y coordinate will be uh when when we have got the final velocity to be equal to 0 right so we have got v equal to or vy will be equal to v0y plus a y t so from the here you can get the time so vy will be equal to 0 this will give you this is 12 minus 2 t right and t is equal to 6 seconds right so at this time you will have the maximum y coordinate right now what you need to do is you need to find out the final velocity vx which is equal to v0x plus a x t right and v0 x is v0 x which is 8 plus acceleration along x is 4 times of 6 which is equal to 32 right meters per second is a velocity of the car and if you see here then at y equal to y max you will have only one velocity that is along the x direction so v will be equal to v x i cap only which is equal to 32 meters per second square sorry second i cap right so this is the answer to your question next b you have to obtain a formula for speed as a function of time so v equal to u plus 80 you can directly apply this so the final speed v will be equal to the initial speed u you is nothing but this 8i plus 12 j these are the initial component of the speeds right plus the accelerations are same you have got 4 i minus 2 j t right so this is the expression of final velocity as a function of time simply you can apply fine or if you want to do the clubbing and then if you take i common then means i let me just show you what i want to do plus if you take j common then this is 12 minus 2 t right so this is what you have and if you want to find out the magnitude of the velocity then it will come out to be root over of 8 plus 40 whole squared plus 12 minus 2 t whole square right so this is the final velocity of final expression of velocity now to find out the time when the car move the fast fastest what you need to do is you need to take v square this is equal to so let's see here we will have 64 plus if we just simplify that so 64 here again 64 t plus 16 t square right plus 144 minus 48 t plus 40 square right and this simplifies to 64 plus 144 4t square right gives you 208 plus 64 minus of 48 16 t plus 20 t squared right now if you see here this is a linear function of time sorry it's a it's an increasing function right because if you see here d v square over d t this is equal to this is a positive time right if you just increase the time then velocity is going to rise right we as time increases.

05:12 So this is 16 plus 20, sorry, 40 t rate.

05:22 And here t is always greater than 0.

05:24 So that means this is an increasing function...

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